查找下一个相同的日历年

给您年份Y，找到下一个与Y相同的日历年。
例子 ：

Input : 2017
Output : 2023

Input : 2018
Output : 2029

如果满足以下两个条件，则年份x与给定的年份y相同。

1. x从y的同一天开始。
2. 如果y是leap年，则x也是。如果y不是leap年，则x也不是。

想法是逐年检查所有年份(从明年开始)。我们跟踪前进的天数。如果总移动天数是7，则当前年份从同一天开始。我们还检查当年的跨度是否与y相同。如果两个条件都满足，我们将返回当前年份。

C++

// C++ program to find next identical year
#include
using namespace std;
 
// Function for finding extra days of year
// more than complete weeks
int extraDays(int y)
{
    // If current year is a leap year, then
    // it number of weekdays move ahead by
    // 2 in terms of weekdays.
    if (y%400==0 || y%100!=0 && y%4==0)
        return 2;
 
    // Else number of weekdays move ahead
    // by 1.
    return 1;
}
 
// Returns next identical year.
int nextYear(int y)
{
    // Find number of days moved ahead by y
    int days = extraDays(y);
 
    // Start from next year
    int x = y + 1;
 
    // Count total number of weekdays
    // moved ahead so far.
    for (int sum=0; ; x++)
    {
        sum = (sum + extraDays(x)) % 7;
 
        // If sum is divisible by 7 and leap-ness
        // of x is same as y, return x.
        if ( sum==0 && (extraDays(x) == days))
            return x;
    }
 
    return x;
}
 
// driver program
int main()
{
    int y = 2018;
    cout << nextYear(y);
    return 0;
}

Java

// Java program to find next identical year
class GFG {
 
// Function for finding extra days of year
// more than complete weeks
static int extraDays(int y)
{
    // If current year is a leap year, then
    // it number of weekdays move ahead by
    // 2 in terms of weekdays.
    if (y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
        return 2;
 
    // Else number of weekdays move ahead
    // by 1.
    return 1;
}
 
// Returns next identical year.
static int nextYear(int y)
{
    // Find number of days moved ahead by y
    int days = extraDays(y);
 
    // Start from next year
    int x = y + 1;
 
    // Count total number of weekdays
    // moved ahead so far.
    for (int sum = 0; ; x++)
    {
        sum = (sum + extraDays(x)) % 7;
 
        // If sum is divisible by 7 and leap-ness
        // of x is same as y, return x.
        if ( sum == 0 && (extraDays(x) == days))
            return x;
    }
 
}
 
// Driver code
public static void main(String[] args)
{
    int y = 2018;
    System.out.println(nextYear(y));
}
}
 
/* This code contributed by PrinciRaj1992 */

C#

// C# program to find next identical year
using System;
     
class GFG
{
 
// Function for finding extra days of year
// more than complete weeks
static int extraDays(int y)
{
    // If current year is a leap year, then
    // it number of weekdays move ahead by
    // 2 in terms of weekdays.
    if (y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
        return 2;
 
    // Else number of weekdays move ahead
    // by 1.
    return 1;
}
 
// Returns next identical year.
static int nextYear(int y)
{
    // Find number of days moved ahead by y
    int days = extraDays(y);
 
    // Start from next year
    int x = y + 1;
 
    // Count total number of weekdays
    // moved ahead so far.
    for (int sum = 0; ; x++)
    {
        sum = (sum + extraDays(x)) % 7;
 
        // If sum is divisible by 7 and leap-ness
        // of x is same as y, return x.
        if ( sum == 0 && (extraDays(x) == days))
            return x;
    }
 
}
 
// Driver code
public static void Main(String[] args)
{
    int y = 2018;
    Console.WriteLine(nextYear(y));
}
}
 
// This code has been contributed by 29AjayKumar

Javascript

 
// JavaScript program for the above approach
 
// Function for finding extra days of year
// more than complete weeks
function extraDays(y)
{
 
    // If current year is a leap year, then
    // it number of weekdays move ahead by
    // 2 in terms of weekdays.
    if (y % 400 == 0 || y % 100 != 0 && y % 4 == 0)
        return 2;
 
    // Else number of weekdays move ahead
    // by 1.
    return 1;
}
 
// Returns next identical year.
function nextYear(y)
{
    // Find number of days moved ahead by y
    let days = extraDays(y);
 
    // Start from next year
    let x = y + 1;
 
    // Count total number of weekdays
    // moved ahead so far.
    for (let sum = 0; ; x++)
    {
        sum = (sum + extraDays(x)) % 7;
 
        // If sum is divisible by 7 and leap-ness
        // of x is same as y, return x.
        if ( sum == 0 && (extraDays(x) == days))
            return x;
    }
 
}
 
// Driver Code
    let y = 2018;
    document.write(nextYear(y));
         
        // This code is contributed by susmitakundugoaldanga.

输出 ：

2029